Question

In the expansion of $(1+x{)}^2(1+y{)}^3(1+z{)}^4(1+w{)}^5,$ the sum of coefficients of the term of degree $12$ is $k.$ Then the value of $\frac{k}{13}$ is

• A. $5$
• B. $6$
• C. $7$
• D. $8$

Answer 1

The correct option is C $7$

Substituting $y=x,z=x,w=x$ in the expression
Coefficient of $x^{12}$ in $(1+x{)}^{14}$ is
${}^{14}{C}_{12}=\frac{14!}{12!\times 2!}=91$
$\{\because {}^n{C}_r=\frac{n!}{r!\times (n-r)!}\}$
$\frac{k}{13}=\frac{91}{13}=7$